Distribution of Residues Modulo p
S. Gun, Florian Luca, P. Rath, B. Sahu and R. Thangadurai
December 11, 2006
1
Introduction
The distribution of quadratic residues and non-residues modulo p has been
of intrigue to the number theorists of the last several decades. Although
Gauss’ celebrated Quadratic Reciprocity Law gives a beautiful criterion to
decide whether a given number is a quadratic residue modulo p or not, it
is still an open problem to find a small upper bound on the least quadratic
non-residue mod p as a function of p, at least when p ≡ 1 (mod 8). This
is because for any given natural number N one can construct many primes
p ≡ 1 (mod 8) having the first N positive integers as quadratic residue (see,
for example, Theorem 3 below).
In 1928, Brauer [1] proved that for any given natural number N one can
find N consecutive quadratic residues as well as N consecutive quadratic
non-residues modulo p for all sufficiently large primes p. Vegh, in a series of
papers ([11], [12], [13] and [14]), studied the distribution of primitive roots
modulo p. He considered problems such as the existence of a consecutive pair
of primitive roots modulo p, or the existence of arbitrarily long arithmetic
progressions of primitive roots modulo p h whose common difference is also
a primitive root mod ph , as well as the existence of a primitive root in a
given sequence of the form g1 + b, g2 + b, · · · , gφ(p−1) + b, where b is any given
integer and the gi ’s are all the primitive roots modulo p.
In 1956, L. Carlitz ([2]) proved that for sufficiently large primes p one
can find arbitrarily long strings of consecutive primitive roots modulo p.
This was independently proved by Szalay ([9] and [10]).
Mathematics Subject Classification: Primary 11N69, Secondary 11A07
Key Words: Quadratic residues, Primitive roots, Finite fields.
1
In [5], some of us studied the problem of the distribution of the nonprimitive roots modulo p. More precisely, we studied the distribution of
the quadratic non-residues which are not primitive roots modulo p. In the
present paper, we improve upon [5] and prove results analogous to those of
Brauer and Szalay. Our main ingredients are some technical results due to
A. Weil [15] or Davenport [4] and Szalay [10].
For convenience, we abbreviate the term ‘quadratic non-residue which is
not a primitive root’ by ‘QNRNP’. Note further that φ(p − 1) = (p − 1)/2
m
if and only if p = 22 + 1 is a Fermat prime. In this case, the set of all
QNRNP’s modulo p is empty, since the primitive roots coincide with the
quadratic non-residues. Thus, throughout this paper we assume that p is
not a Fermat prime. We prove the following theorems.
Theorem 1. Let ε ∈ (0, 1/2) be fixed and let N be any positive integer.
Then for all primes p ≥ exp((2ε−1 )8N ) satisfying
1
φ(p − 1)
≤ − ε,
p−1
2
we can find N consecutive QNRNP’s modulo p.
Theorem 1 above generalizes the results of A. Brauer [1] and S. Gun, et
al. [5].
Given a prime number p, we let
k :=
p−1
− φ(p − 1)
2
denote the number of QNRNP’s modulo p and we write g 1 < g2 < . . . < gk
for the increasing sequence of QNRNP’s.
Corollary 1. For any given ε ∈ (0, 1/2) and natural number N , for all
primes p ≥ exp((2ε−1 )8N ) and satisfying φ(p − 1)/(p − 1) ≤ 1/2 − ε, the
sequence g1 + N, g2 + N, . . . , gk + N contains at least one QNRNP.
Theorem 2. There exists an absolute constant
that for almost
c 0 > 0 such
log p
all primes p, there exist a string of N p = c0
of quadratic nonlog log p
residues which are not primitive roots.
We may also combine our Theorems with above mentioned results of
Brauer and Szalay and infer that if ε ∈ (0, 1/2) and N are fixed, then for
each sufficiently large prime p with φ(p − 1)/(p − 1) < 1/2 − ε, there exist
2
N consecutive quadratic residues, N consecutive primitive roots, as well as
N consecutive quadratic non-residues which furthermore are not primitive
roots. In fact, we can even arrange the quadratic residues to be the first N
quadratic residues.
Theorem 3. For every positive integer N there are infinitely many primes
p for which 1, 2, . . . , N are quadratic residues modulo p, and there exist both a string of N consecutive QN RN P ’s as well as a string of N
consecutive primitive roots. The smallest such prime can be chosen to be
< exp(exp(c1 N 2 )), where c1 > 0 is an absolute constant.
2
Preliminaries
Unless otherwise specified, p denotes a sufficiently large prime number. We
denote the group of residues modulo p by Z p and the multiplicative group
of Zp by Z∗p .
An element ζ ∈ Z∗p is said to be a primitive root modulo p if ζ is a
generator of Z∗p . Once we know a primitive root modulo p, the QNRNP’s
are precisely the elements of the set
n
o
ζ ` : ` = 1, 3, . . . , (p − 2) and (`, p − 1) > 1 .
Consider a non-principal character χ : Z ∗p −→ µp−1 , where µp−1 denotes the
group of (p − 1)th roots of unity. Then it is easy to observe that χ(ζ) is a
primitive (p − 1)th root of unity if and only if ζ is a primitive root mod p.
Let η be a primitive (p − 1)th root of unity and assume that χ(ζ) = η. Since
χ is a homomorphism, it follows that χ(ζ i ) = χi (ζ) = η i . Hence, by the
above observation, it is clear that χ(κ) = η i with (i, p − 1) > 1 with some
odd i if and only if κ is a QNRNP mod p.
Let ` be any non-negative integer. We define
X
`
β` (p − 1) =
ηi .
1≤i≤p−1
i odd, (i,p−1)>1
Lemma 1. For 0 < l < p − 1, we have
β` (p − 1) = −α` (p − 1),
where α` (p − 1) is the sum of the `th powers of the primitive (p − 1)th roots
of unity.
3
Proof. Observing that
p−2
X
(p−3)/2
i
η =0=
i=0
X
η 2i ,
i=0
we get the desired result.
Let
χ2 = χ21 ,
χ1 ,
χp−2 = χp−2
1 ,
...,
χ0 = χp−1
1 ,
be all the multiplicative characters modulo p with the convention χ ` (0) = 0
for all ` = 0, 1, . . . , p − 2.
Lemma 2. We have,
p−2
X
`=0
β` (p − 1)χ` (x) =
p − 1,
0,
if x is a QNRNP;
otherwise.
Proof. When x ≡ 0 (mod p), the statement is obvious. We assume that
x 6≡ 0 (mod p). Let η be a primitive (p − 1)th root of unity. Consider
η i1 , η i2 , . . . , η ik ,
(ij , p − 1) > 1
where
1 < i 1 < . . . < ik ,
and
and ij is odd for all j = 1, 2, . . . , k.
The expression
1 + η il χ1 (x) + η il
2
χ2 (x) + . . . + η il
p−2
χp−2 (x)
has the value p − 1 if (χ1 (x))−1 = η il and zero otherwise whenever x 6= 0.
Thus, giving l the values 1, 2, . . . , k, and adding up the above resulting
expressions we get
p − 1, if x is QNRNP;
β0 (p − 1)χ0 (x) + . . . + βp−2 (p − 1)χp−2 (x) =
0,
otherwise,
which completes the proof of the lemma.
The following deep theorem of A. Weil [15] is of central importance in
the proofs of Theorem 1 and Theorem 2.
4
Theorem 4. For any integer ` satisfying 2 ≤ ` < p and for any nonprincipal characters χ1 , χ2 , · · · , χ` and distinct a1 , a2 , . . . , a` ∈ Zp , we have
p
X
√
χ1 (x + a1 )χ2 (x + a2 ) · · · χ` (x + a` ) ≤ (` − 1) p.
x=1
For ` = 2, Davenport [3] was the first one to prove the above bound.
Note also that when ` = 1, the sum is 0.
For a positive integer m, we write ω(m) for the number of distinct prime
factors of m. The next result is due to Szalay [9].
Lemma 3. We have,
p−2
X
`=0
3
|α` (p − 1)| = 2ω(p−1) φ(p − 1).
The Proof of Theorem 1
Let M (p, N ) denote the number of consecutive QNRNP modulo p of length
N in Z∗p . We shall start with the following technical lemma.
Lemma 4. For any prime p and any positive integer N , we have
N k
√
≤ 2N 2N ω(p−1) p.
M (p, N ) − p
p−1
Proof. First note that β0 (p − 1) = k. Clearly, by Lemma 2, we have

"
#
p−N
p−2
−1

X NY
1 X
M (p, N ) =
β` (p − 1)χ` (x + j)


p−1
x=1
j=0
`=0


"
#
p N
p−2
−1

X
Y
1 X
=
β` (p − 1)χ` (x + j)


p−1
x=1
j=0
`=0

"
#
p N
p−2
−1

X
Y
X
= (p − 1)−N
k+
β` (p − 1)χ` (x + j)


x=1
= p
k
p−1
N
+
j=0
A
,
(p − 1)N
5
`=1
where
X
A=
0≤l1 ,l2 ,··· ,lN ≤p−2
(l1 ,...,lN )6=0


N
Y
j=1



p
N
X
Y

βlj (p − 1)
χlj (x + j − 1) .
x=1
j=1
In order to finish the proof of Lemma 4, we have to estimate A. So, we
rewrite it as A = B + C, where




p
N
N
Y
X
X
Y


χlj (x + j − 1) ,
C=
βlj (p − 1)
1≤l1 ,l2 ,··· ,lN ≤p−2
x=1
j=1
j=1
and B is the similar summation with at least one (but not all) of the l j ’s
equal to zero. We further separate each sum over the set for which exactly
one `i ’s is zero, then exactly two of the `i ’s are 0, etc., up to when just one
of the `i ’s is nonzero.
Now, we look at the sum corresponding to the case when exactly j of
the `i ’s are equal to zero. This means that N − j of the ` i ’s are non-zero.
The corresponding sum is
"N −j
# " p N −j
!
#
X
Y
X Y
βrb (p − 1)
χrb (x + mb ) + E ,
Bj = k j
0<r1 ,...,rN −j ≤p−2
x=1
b=1
b=1
where E is the sum of some (p − 1)th roots of unity and in the summation
at most N terms occur. When we take the absolute value of this summand,
we get
p N −j
!
!
N
−j
X Y
Y
X
j
χrb (x + mb ) + N
|βrb (p − 1)| |Bj | ≤ k
0<r1 ,...,rN −j ≤p−2
≤k
j
p−2
X
`=0
|β` (p − 1)|
x=1
b=1
!N −j
p
X
x=1
N
−j
Y
b=1
b=1
!
!
χrb (x + mb ) + N .
Now, note that |β` (p − 1)| = |α` (p − 1)| for all ` = 1, 2, . . . , p − 2, and
|β0 (p − 1)| = k, while |α0 (p − 1)| = φ(p − 1). Thus, by Theorem 4 and
Lemma 3, we get
N −j
√
|Bj | < k j 2ω(p−1) φ(p − 1)
((N − j − 1) p + N )
N −j √
< 2N k j 2ω(p−1) φ(p − 1)
p.
6
This inequality holds for all j = 1, 2, . . . , N − 2. When j = N − 1, we get
|BN −1 | ≤ k N −1 2ω(p−1) φ(p − 1)N.
The term C in A can also be estimated as above and we get for it
N
√
(N − 1) p.
|C| ≤ 2ω(p−1) φ(p − 1)
So, we see that the inequality (1) holds when j = N − 1 as well. Adding up
all the above estimates for |Bj | and |C|, we get
A
(p − 1)N
√
N
−1 N −j
X
p
N j ω(p−1)
≤ 2N
2
φ(p
−
1)
k
(p − 1)N
j
j=0
N
k
√
ω(p−1) φ(p − 1)
+
< 2N p 2
p−1
p−1
N ω(p−1) √
p,
< 2N 2
where we used the fact that 2ω(p−1) φ(p − 1)/(p − 1) + k/(p − 1) < 2ω(p−1) ,
which finishes the proof of the lemma.
Proof of Theorem 1. We assume that N ≥ 4. From the definition of k, it is
easy to observe that
1 φ(p − 1)
k
= −
≥ ε.
p−1
2
p−1
Lemma 4 above tells us now that
N k
√
pεN − M (p, N ) ≤ M (p, N ) − p
≤ 2N 2N ω(p−1) p,
p−1
The above chain of inequalities obviously implies that M (p, N ) > 0 if
√ N
pε > 2N 2N ω(p−1) .
(1)
This last inequality is fulfilled if
log p > 2 log(2N ) + 2N ω(p − 1) log 2 + log(ε−1 ) .
(2)
For p > 4 · 106 , we have that ω(p − 1) < 2 log p/ log log p. Thus, for such
values of p, the right hand side above is bounded above by
2 log(2N ) +
4N log 2
log p + 2N log(ε−1 ),
log log p
7
and so the desired inequality is fulfilled provided that
4N log 2
1−
log p > 2 log(2N ) + 2N log(ε−1 ).
log log p
When p > exp(28N ), the factor appearing in parenthesis in the left hand
side of the last inequality above is ≥ 1/2. Note that since N ≥ 1, we have
that exp(28N ) > 4 · 106 , so the inequality ω(p − 1) < 2 log p/(log log p) is
indeed satisfied for such values of p. Thus, in this range for p it suffices that
log p ≥ 4 log(2N ) + 4N log(ε−1 ),
leading to p ≥ (2N )4 ε−4N . Since (2N )4 ≤ 24N , the inequality
exp((2ε−1 )8N ) > max{exp(28N ), (2N )4 (ε−1 )4N }
holds for all ε ≤ 1/2 and N ≥ 1, so the proof of Theorem 1 is completed.
4
The Proof of Theorem 2
Let P be the set of all primes. Fix δ > 0 and let P 1 be the set of all primes
p ∈ P such that |ω(p − 1) − log log p| < δ log log p and p − 1 is divisible
by some odd prime q ≤ log log p. It is well-known that P 1 contains most
primes; that is, if x is large then the set of primes p ∈ P\P 1 is of cardinality
o(π(x)) as x → ∞.
We now let x be a large positive real number. Let p ≤ x be a prime. We
assume that p > x/ log x, since there are only π(x/ log x) = o(π(x)) primes
p ≤ x/ log x. Then log p ≥ log x − log log x, so log log p = log log x + O(1).
Thus, if p ∈ P1 ∩[x/ log x, x] and x is large, then ω(p−1) ≤ (1+2δ) log log x.
Furthermore, if q is the smallest odd prime factor of p−1, then φ(p−1)/(p−
1) ≤ 1/2 − 1/(2q), and since 2q ≤ 2 log log x, we can take ε = 1/(2 log log x)
and hence ε−1 = 2 log log x. With all these choices, inequality (2) will be
fulfilled if
log x − log log x > 2 log(2N )
+ 2N ((1 + 2δ) log log x log 2 + log(2 log log x)) .
log x
, where we
The above inequality is satisfied if we choose N = c3
log log x
can take c3 to be a positive constant < 1/(2 log 2), provided that afterwards
δ is chosen to be small enough and x is then chosen to be sufficiently large,
which completes the proof of this theorem.
8
5
Proof of Theorem 3
Proof of Theorem 3. First we prove that there exist infinitely many
primes p for which 1, 2, . . . , N are all quadratic residues modulo p for any
given natural number N . For each prime q ≥ 5 let a q (mod q) be a quadratic
residue modulo q such that aq > 1 and put a3 = 1. Let p be a prime
congruent to 1 modulo 8 and to aq modulo q for all odd primes q ≤ N .
Then, by Quadratic Reciprocity,
p
aq
q
=
=
=1
p
q
q
2
= 1 because p ≡ 1
whenever q ≤ N is an odd prime. Furthermore,
p
(mod 8). Using
the multiplicativity property of the Legendre symbol, we
a
get that
= 1, whenever a is a positive integer all whose prime factors
p
are ≤ N . In particular, the first N positive integers are quadratic residues
modulo p. Note that 3 | (p − 1), and from the argument used at the proof
of Theorem 2, it follows that we may take ε = 1/6. Furthermore, p − 1
is not divisible by any prime q ∈ [5, . . . , N ]. By the Chinese Remainder
Theorem, the system of congruences p ≡ 1 (mod 8) and p ≡ a q (mod
Q q) for
all odd primes q ≤ N has a solution p0 (mod P ), where P = 4 q≤N q =
exp(O(N )). There are infinitely many primes in this progression. Now the
argument from the proof of Theorem 1 shows that such p can be chosen
on the scale of x = exp(128N ). The only problem that might worry us is
the existence of primes in the arithmetic progression p 0 (mod P ) on the
scale of x. But note that P = exp(O(N )) = (log x) o(1) , so the SiegelWalfitz Theorem, for example, tells us that the interval [x, 2x] contains
(1 + o(1))π(x)/φ(P ) primes p ≡ p0 (mod P ) (in particular, at least one of
them), which finishes the argument.
6
Final Remarks
Let N 6= 1 be any square-free natural number. Then it is well-known that N
is a quadratic non-residue modulo p for infinitely many primes p. The analogous result for primitive roots is known as Artin’s Primitive Root conjecture.
In 1967, Hooley [7] proved this conjecture subject to the assumption of the
generalized Riemann hypothesis. Interestingly, it is not even known whether
2 is a primitive root modulo infinitely many primes. For more details, we
9
refer to the article by Ram Murty [8]. Finally, in Theorem 1, it would be of
interest to obtain a constant M which depends only on the natural number
N and not on ε.
Acknowledgments. We are grateful to Prof. Ram Murty for going through
our work. This work was started when the first and the third authors were
at Harish-Chandra Research Institute. The second author was supported in
part by grants PAPIIT IN104505, SEP-CONACyT 46755 and a Guggenheim
Fellowship.
References
[1] A. Brauer, Űber Sequenzen von Potenzresten, Sitzungsberichte der
Preubischen Akademie der Wissenschaften, (1928), 9-16.
[2] L. Carlitz, Sets of primitive roots, Compositio Math., 13 (1956), 65-70.
[3] H. Davenport, On the distribution of the l-th power residues mod p, J.
London Math. Soc., 7 (1932), 117-121.
[4] H. Davenport, On character sums in finite fields, Acta Math., 71 (1939),
99-121.
[5] S. Gun, B. Ramakrishnan, B. Sahu and R. Thangadurai, Distribution
of Quadratic non-residues which are not primitive roots, Math. Bohem.,
130 (2005), no. 4, 387-396.
[6] G. H. Hardy and E. M. Wright, “An Introduction to the Theory of
Numbers,” 4th ed. Clarendon, Oxford, 1960.
[7] C. Hooley, On Artin’s conjecture, J. Reine Angew. Math 226 (1967)
209-220.
[8] M. R. Murty, Artin’s conjecture for primitive roots, Math. Intelligencer,
10 (1988), no. 4, 59-67.
[9] M. Szalay, On the distribution of the primitive roots mod p (in Hungarian), Mat. Lapok, 21 (1970), 357-362.
[10] M. Szalay, On the distribution of the primitive roots of a prime, J.
Number Theory, 7 (1975), 183-188.
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[11] E. Vegh, Primitive roots modulo a prime as consecutive terms of an
arithmetic progression, J. Reine Angew. Math., 235 (1969), 185-188.
[12] E. Vegh, Primitive roots modulo a prime as consecutive terms of an
arithmetic progression - II, J. Reine Angew. Math., 244 (1970), 108111.
[13] E. Vegh, A note on the distribution of the primitive roots of a prime,
J. Number Theory, 3 (1971), 13-18.
[14] E. Vegh, Primitive roots modulo a prime as consecutive terms of an
arithmetic progression - III, J. Reine Angew. Math., 256 (1972), 130137.
[15] A. Weil, On the Riemann hypothesis, Proc. Nat. Acad. Sci. U.S.A, Vol.
27 (1941), 345-347.
Addresses of the Authors
S. Gun
Florian Luca
Department of Mathematical and
Instituto de Matemáticas
Computational Sciences
Universidad Nacional Autónoma de México
3359 Mississauga Road
C.P. 58089, Morelia, Michoacán
North Mississauga,
México
ON, Canada, L5L 1C6
[email protected]
[email protected]
P. Rath
Institute of Mathematical Sciences
C. I. T. Campus, Taramani
Chennai - 600113, INDIA
[email protected]
B. Sahu
Harish-Chandra Research Institute
Chhatnag Road, Jhunsi
Allahabad 211019, INDIA
[email protected]
R. Thangadurai
Harish-Chandra Research Institute
Chhatnag Road, Jhunsi
Allahabad 211019, INDIA
[email protected]
11